I am solving a calculus of variations problem. I am trying to show that given a piecewise smooth real-valued function $x$ on $[1,2]$ with $x(1) = 1$ and with no information on $x(2)$:
$$\int_2^1 t^2x^\prime(t)^2 + 2x(t)^2 dt + 4x(2)^2 \geq 2$$
For clarity, $x$ being piecewise smooth on $[1,2]$ means that $x$ is continuous on $[1,2]$, has a finite number of corner points on $[1,2]$, $x$ is $C^1$ away from those corner points, and at $1$, $2$, and any corner points, the left- and/or right-hand derivatives of $x$ exist (but of course may not be equal).
Note that I do not actually know whether this is true but I think that this is true.
Here is the best attempt I made:
Since $t \in [1,2]$:
$$\int_2^1 t^2x^\prime(t)^2 + 2x(t)^2 dt + 4x(2)^2 \geq \int_1^2 x^\prime(t)^2 + 2x(t)^2 dt + 4x(2)^2$$
Then:
$$\int_1^2 x^\prime(t)^2 + 2x(t)^2 dt + 4x(2)^2 = \int_1^2 x(t)^2 dt + \int_1^2 x^\prime(t)^2 + x(t)^2 dt + 4x(2)^2$$
Applying the AM-GM inequality to the second term:
$$\int_1^2 x(t)^2 dt + \int_1^2 x^\prime(t)^2 + x(t)^2 dt + 4x(2)^2 \geq \int_1^2 x(t)^2 dt + \int_1^2 2\sqrt{x^\prime(t)^2x(t)^2} dt + 4x(2)^2$$
$$= \int_1^2 x(t)^2 dt + 2\int_1^2 |x^\prime(t)||x(t)| dt + 4x(2)^2 \geq \int_1^2 x(t)^2 dt + 2\left|\int_1^2 x^\prime(t)x(t) dt\right| + 4x(2)^2$$
$$= \int_1^2 x(t)^2 dt + 2\left|[x(t)^2]_1^2 - \int_1^2 x^\prime(t)^2 dt\right| + 4x(2)^2$$
but now I am stuck with the pesky minus sign in the absolute value.
I think I found a solution, but anyone feel free to comment if you do not think this is correct.
We make two observations. First, using integration by parts:
$$\int_1^2 x(t)x^\prime(t) dt = [x(t)^2]_1^2 - \int_1^2 x^\prime(t)^2 dt$$
and rearranging:
$$\int_1^2 x(t)x^\prime(t) dt + \int_1^2 x^\prime(t)^2 dt = [x(t)^2]_1^2 = x(2)^2 - x(1)^2$$
Since $\displaystyle \int_1^2 x^\prime(t)^2 dt \geq 0$:
$$\int_1^2 x(t)x^\prime(t) dt \leq x(2)^2 - x(1)^2$$
and thus:
$$-\int_1^2 x(t)x^\prime(t) dt \geq x(1)^2 - x(2)^2$$
Secondly, recall that for $a,b \in \mathbb{R}$, $(a+b)^2 = a^2 + b^2 + 2ab$ and so $a^2 + b^2 = (a+b)^2 - 2ab$ and thus $a^2 + b^2 \geq -2ab$.
Hence, continuing from above:
$$\int_1^2 x(t)^2 dt + \int_1^2 x^\prime(t)^2 + x(t)^2 dt + 4x(2)^2 \geq \int_1^2 x(t)^2 dt - 2\int_1^2 x(t)x^\prime(t) dt + 4x(2)^2$$
$$\geq \int_1^2 x(t)^2 dt + 2(x(1)^2 - x(2)^2) + 4x(2)^2 = \int_1^2 x(t)^2 dt + 2x(1)^2 + 2x(2)^2 \geq 2x(1)^2 = 2$$